The concept of probability is very important for managers to make judicious decisions. The applications of this concept can be found in numerous areas like insurance, stock markets, medical diagnostics, Quality testing to name a few. Learning this concept in a systematic way improves our decision making abilities apart from scoring good marks in exams. Read every definition and try to understand even though your instinct easily get the answer with out these concepts. This understanding helps you get solutions for some tough problems later.
Important concepts and formulas:
Experiment: Phenomenon where outcomes are un-certain. Single throw of a six-sided die is an experiment
Sample space**: Set of all outcomes of the experiment. {1,2, 3,4, 5, 6} are the total outcomes of rolling a six faced dice.
Event**: A collection of outcomes; a subset of S Example: An event of occurrence of even number is {2,4,6}
Note: You must remember these two starred definitions
Note: You must remember these two starred definitions
Equally likely Events: If
each outcome of an experiment has equal chance of occurrence then we
say, the outcomes are equally likely. Getting heads or tail are equally
likely.
Exhaustive Events: A set
of events is jointly or collectively exhaustive if at least one of the
events must occur. For example, when rolling a six-sided die, the
outcomes 1, 2, 3, 4, 5, and 6 are collectively exhaustive, because they
encompass the entire range of possible outcomes.
Another
way to describe collectively exhaustive events, is that their union
must cover all the events within the entire sample space. For example,
events A and B are said to be collectively exhaustive if (A∪B=S)
Probability: The ratio between the required outcomes to the total possible outcomes of an experiment.
Probability = Required OutcomesPossible Outcomes
Types of probabilities:
Union probability: The probability of either of the two events is called as union probability.
If A and B are two events, then Union probability isP(A∪B)
If A and B are two events, then Union probability is
Example: What is the probability of getting a king or a diamond from a well shuffled pack of 52 cards.
Say, A is the event of drawing a king and B is the event of drawing a diamond and A∩B
is drawing a diamond king and
is drawing either diamond or king.
we know that P(A∪B)=P(A)+P(B)−P(A∩B)
Intersection probability: The
probability of the occurrence of both the events is called intersection
probability. If A and B are two events, then intersection probability
is P(A∩B)
Example: What is the probability of getting a black king and red card from a well shuffled pack of 52 cards with replacement.
We
know that a suit contains four kings and 26 red cards. Say, P(A) is
drawing a king and P(B|A) is drawing a red card given we have drawn a
king in the first event. So drawing a king from a suit ⇒P(A)=452=113
But this first event does not have any affect on the second event. so P(B|A) = P(B)
Complement probability: The complement of a sample set A is denoted by AC
.
Example: What is the probability of getting an even number when rolling a dice.
Let A is the event of getting an odd number. So A = {1,3,5}, then Ac
= {2,4,6}
Then P(Ac)
= 1/3
Conditional probability: The probability of an event A, given event B has already happened is given by conditional probability.
means, The probability of A, if event B has already happend.
means, The probability of B, if event A has already happend.
From above P(A∩
B) = P(A)
P(A | B)
Example:
On rolling a dice, what is the probability of getting 1, given that the number is odd.
Let
the probability of getting 1 is P(A), the probability of getting an odd
number is P(B) and the probability of getting 1 given the number is odd
P(A|B)
. We know that A = {1}, B= {1,3,5} So
Mutually exclusive and Independent Events:
Mutually exclusive events: If two events are mutually exclusive, the probability of occurrence of the two events at the same time is
zero. Or the happenning of one event prevents all other events
from happenning. For example, the probability of getting both 1 and 6
with a single rolling of dice is
impossible. So occurrence of 1 and occurence of 6 both are mutually
exclusive events. (Have you seen, Exclusive livecast of the cricket
match prevents other channels broadcasting the same cricket match!!)
If two events are mutually exclusive, P(A∩B)
= 0; and
Independent events: It
is a common practice to consider independent experiments and
independent events are same. For example, the outcomes are independent
when we roll a dice for two times. The result of the first experiment
may not effect the outcome of the second experiment.
From the definition, we know that an event is a subset of the sample space S. Say A and B are two events.
If two events are independent then P(A)=P(A|B)
or
From conditional probability formula P(A|B)=P(A∩B)P(B)
But P(A)=P(A|B)
So P(A∩B)=P(A)×P(B)
Example: An
event A is defined as getting either 1 or 2 on rolling a dice. If An
event B is getting an odd number on rolling the dice, Then what is the
probaility of event A given B has occurred.
A = {1, 2}
B = {1, 3, 5}
= {1}
P(B) = 3/ 6 = 1/2
= 1/6
But P(A) = 2/6 = 1/3
This
is interesting. Events A and B are independent as the additional
information that B is odd does not change the probability of A.
If two events are independent then P(A∩B)=P(A)×P(B)
Relation between Mutually exclusive and independent Events:
Firstly, never bother about the two above terms. But remember if
two events are mutually exclusive then they are not independent. and if
two events are independent they cannot be mutually exclusive.
We know that P(A|B)=P(A∩B)P(B)
If two events mutually exclusive P(A∩B)
= 0 so
and For A and B are independent P(A|B) = P(A)
and P(A).P(B) = 0, which is not possible.
so there must be common event.
Practice Problems (Level - 1)
1. If a fair dice is thrown twice, what is the probability of getting at least one six?
We
can apply complement probability rule to solve this problem easily.
Assumed P(A) is the probability of getting atleast one six. then P(AC)
is not getting a single six. A = {6} and
= {1, 2, 3, 4, , 5} and S = {1, 2, 3, 4, 5, 6}. On throwing a dice two
times the events are independent. So the probability of not getting a
six in the first throw is 5/6 and in the second throw is 5/6 as these
two events are independent.
So P(AC)
=
we know that P(A) + P(AC)
= 1 so P(A) = 1- 25/36 = 11/36
Alternate method:
We
can count the points in the event A as {(1,6), (2, 6), (3, 6), (4, 6),
(5, 6), (6, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)} = 11
S = 36
P(A) = 11/36
2. There are two positive integers a and b. What is the probability that a + b is odd?
The sample space S for adding two numbers is Even + Even, Even + Odd, Odd + Even, Odd + Odd
If a + b has to be odd, then Even + Odd, Odd + Even are the points in the event. So Probability = 2/4 = 1/2
3.
A number is selected at random from the number 1, 2, ...,50. What is
the probability that the number is multiple of either 6 or 9?
Say the event A is getting multiples of 6, and the event B is getting multiples of 9.
A = {6, 12, 18, .....48} = 8 ⇒P(A)=850
B = {9, 18, ...........45} = 5 ⇒P(B)=550
But
events A and B have some common points 18, 36. This can be calculated
by taking LCM of 6, 9 and finding the multiples of the LCM below 50.
So A∩B
= 2
From union probability formula P(A∪B)=P(A)+P(B)−P(A∩B)
4.
There are 6 men and 7 women in a committee. 2 people need to be
selected as representatives of this committee. What is the probability,
that out of these 2 people, 1 is a man and another is a woman?
Number of ways of selecting 2 persons out of 13 persons is 13C2
Number of ways of selecting 1 man out of 6 is 6C1
Number of ways of selecting 1 woman out of 6 is 7C1
As the above two events are independent, the number of ways of selecting one man and one women = 6C1×7C1
So the probability of selecting one man and one women from a group of 13 = 6C1×7C113C2
5. Anil’s
age is currently between 15-25 years (exclusive of 15, 25). What is the
probability that after 15 years, he is 37 years or above in age?
Anil's
age after 15 years is in between 30 and 40 exclusive of 30, 40. Number
of points in the sample space are {31, 32, ........39}= 9 and number of
points in the event are {37, 38, 39} = 3
So required probability = 3/9 = 1/3
6. The probability that A, B and C will live for more than 60 years is 12,13and14
respectively. What is the probability that atleast 1 of them will be alive after 60 yrs of age?
We use complement probability to solve this question easily. Let A is the event of atleast one of them will be alive. Then AC
is the probability that nobody lives.
As the three persons lifes are independent of each other, we use product rule to get total probability
The probability of A dies = 1−12=12
The probability of B dies = 1−13=23
The probability of C dies = 1−14=34
So the probability that all of them die = 12×23×34=14
The probability that atleast one person will be alive = 1 - 14
=
7. In the last question, what is the probability that atleast 2 are alive?
Any 2 are alive = (A + B) alive and C dead OR (B + C) alive and A dead OR (A + C) alive and B dead. Let A¯¯¯¯
is the probability A dies. The the probability of atleast two will be alive will be given as
From above we have already calculated the probabilities of A, B, C not alive and the events are independent.
So by substituting in the above equation and by applying the product rule 12×13×14+12×23
8.
What is the probability that when 3 cards are pulled from a pack of
cards, without replacement, that we get 1 King, 1 Queen and 1 Jack?
Here the order of picking up is not specified and the events are dependent. Total possibility of picking up the given cards are
1st Pick 2nd Pick 3rd Pick
K Q J
K J Q
Q K J
Q J K
J Q K
J K Q
The probability of Picking up a king in the first attempt = 4/52
The probability of Picking up a queen in the second attempt = 4/51
The probability of Picking up a Jack in the third attempt = 4/50
and total probability = (452×451×450)×6=165525
Conditional Probability:
9. A bag contains 5 black and 3 red balls. A ball is taken out of the bag and is not replaced.
(a) What is the probability of drawing a red ball in the second draw if the ball in the first draw is red?
(b) What is the probability of drawing a black ball in the second draw if the ball in the first draw is red?
Sol:
A: Drawing red in 1st draw
B: Drawing red in 2nd draw
C: Drawing black in 2nd draw
(a)
If one ball is drawn already and it is a red ball, the remaining balls
are 7 and only 2 red balls left in the bag. So P(B|A) = 27
(b) Similarly P(C|A) = 57
Alternatively:
n(A) =RR + RB = 3C1.2C1+3C1.5C1=21
= 6
a. P(B|A)
=
b. P(C|A)
=
10. A coin is flipped twice. What is the probability that both flips result in heads given that first flip does?
A = First flip lands heads
B = Second flips is head
11. A family has two children. What is the probability that both boys are boys given that one of the children is boy?
Sol: A = One of the children is boy = (BB, BG, GB)
B = Second one is boy = (BB)
Probability = 1/3
12.
Ten cards numbered 1 to 10 are placed in a box. And one card is drawn
randomly. If is it is known that the drawn card is more than 3, what is
the probability that it is an even number?
A = Drawn card is more than 3 = (4, 5, 6, 7, 8, 9, 10) ⇒
n(A) = 7
B = It is an even number = (2, 4, 6, 8, 10)
P(B | A) = 4 / 7
Multiplication Theorem:
13.
Suppose that an Urn contains 8 red balls and 4 white balls. We draw 2
balls without replacement. What is the probability that both balls are
red?
Sol: P(R1∩R2)=8C212C2=1433
Alternatively: P(R1∩R2)=P(R1)×P(R2|R1)
=
14.
A bag contains 15 items, of which 4 are defective. The items are
selected at random, not put back. What is the probability that 10th one
examined is last defective?
In this problem, we should get 3 defects in the first 9 examination and last defect one should be 10th one.
A = {Exactly 3 defects in 9 examined}
B = { 10th one is defective}
Once 3 defects are drawn in 9 examinations, last defect should be in 10th one. So this one is 1 out of remaining 6.
So P(B|A)=16
Other models:
15.
Three groups of children contain 3 girls and one boy; 2 girls and 2
boys, one girl and 3 boys. One child is selected at random from each
group. What is the probability that the three selected consists of 1
girl and 2 boys?
1 girls and 2 boys can be selected as GBB, BGB, BBG from each of the three given groups.
GBB: 34×24×34=932
BGB: 14×24×34=332
BBG: 14×24×14=132
Since above three are mutually exclusive cases, therefore the required probability = 932+332+132=1332
16.
A student is given a true - false exam with 10 questions. If he gets 8
or more correct answers he passes the exam. Given that he guesses at
the answer to each question, what is the probability that he passes the
exam.
As this exam is a true false exam, every question has two answers. So total ways of answering the paper = 2×2×..........2(10times)=210
= 1024
Now he can pass the exam, if his correct answers are 8, 9 or 10.
Number of ways of answering 8 questions correctly = 10C8
(Think that there are 10 blanks where we need to put either T or F. Selecting 8 places to put 8 T's is
) = 45
Number of ways of answering 9 questions correctly = 10C9
= 10
Number of ways of answering 10 questions correctly = 10C10
= 1
Total possible ways of getting 8, 9 or 10 questions correctly is = 56
So the probability to pass the exam = 561024=7128
17.
A Bag contains 2 white and 2 black balls. A ball is drawn at random.
If it is white, it is not replaced into the urn, otherwise it is
replaced along with another ball of the same color. The process is
repeated. Find the probability that the third ball drawn is black
The first ball can be drawn in the following for possible ways
1. Both the first and the second balls drawn are white (2B)
2. The first ball drawn is white adn the second ball drawn is black (1W + 3B)
3. The first ball drawn is black and the second ball drawn is white (1W + 3B)
4. Both the first and the second balls drawn are black (2W + 4B)
The values in the brackets indicates the number of balls after two ball have drawn and made necessary changes required.
Let the above 4 cases are denoted by E1,E2,E3andE4
respectively. Also E denote the event that the third ball drawn is black.
Now P(E1)
=
Now P(E|E1)
=
has happened
=
18. A and B roll a pair of dice. A wins if he roll 6 and B wins if he rolls 7. They continue starting with A till one of them throws their respective number. What is the probability that A wins?
Let
There are 5 ways of obtaining 6 {(1,5), (5, 1), (2, 4), (4, 2), (3, 3)}
and Sample space while throwing 2 dice =
So
It is given that A starts the game and he will win in the following mutually exclusive ways.
(1)
(2)
(3)
.....................
Hence the probability of A winning say P(A) is given by
P(A) =
=
=
Taking
=
Above is in the Geometric progression with common difference of
Formula for infinite geometric progression is =
Here 'a' is the starting term, and r is the common difference.
=